Combinatorics problem?

COMBINATORICS has 13 letters, distributed as follows:

one A

one B

two Cs

two Is

one M

one N

two Os

one R

one S

one T

There are 10 distinct letters.

How many ways are there to form a list of two distinct letters from the set of letters

in the word COMBINATORICS?

Taking "ways to form a list" as referring to permutations" (that is, the same letters in different order are a different "way"), we have

10 * 9 = 90 ways.

[If you need any help understanding permutations or combinations formulas, I recommend this site:

http://www.mathsisfun.com/combinatorics/combinatio...

for both clarity and completeness.]

three distinct letters?

10 * 9 * 8 = 720 ways

four distinct letters?

10 * 9 * 8 * 7 = 5,040 ways

How many ways are there to form a list of two letters from the set of letters in

the word COMBINATORICS if the letters cannot be used more often than they

appear in COMBINATORICS?

There are the 90 ways we had for lists of 2 distinct letters, plus 3 lists in which the same letter appears twice (CC, II, and OO), for a total of 93 ways.

three letters?

We can start with the 720 ways we found for lists of 3 distinct letters, and add in the number of lists in which a letter is repeated.

There are 3 letters which can appear twice (C, I, and O), and none that can appear 3 times.

Each of the 3 repeatable letters can be combined with any of 9 other letters appearing once.

The letter that appears only once can appear in 3 different places in the list.

Thus, there are

3 * 9 * 3 = 81 lists with a letter repeated in them,

and a total of

720 + 81 = 801 ways to make three-letter lists under these rules.